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fiasKO [112]
3 years ago
10

What forms the organic portion of soil?

Chemistry
1 answer:
telo118 [61]3 years ago
5 0

Answer:

<h2>Dead & Decaying Organisms</h2>

Explanation:

There are many inorganic components of soil, such as clay. However, soil is mainly comprised of dead and decaying organisms. For example, old, and decaying plants.

<em>PLEASE MARK BRAINLIEST</em>

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The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p
almond37 [142]

Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen               40300 g

Carbon                11160 g

Hydrogen            6200 g

Nitrogen              1860 g

Calcium                992 g

Phosphorus         744 g

Other elements   744 g

It is possible to check it adding all the components, the total is 62000 g.  

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

                            Percentage ( %)

Oxygen               65.0

Carbon               18.0

Hydrogen            10.0

Nitrogen              3.0

Calcium               1.6

Phosphorus         1.2

Other elements   1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

                           Percentage ( %)      Decimal number

Oxygen               65.0                         65/100 = 0.65

Carbon               18.0                           18/100 = 0.18

Hydrogen            10.0                         10/100 = 0.10

Nitrogen              3.0                           3/100 = 0.03

Calcium               1.6                           1.6/100 = 0.016

Phosphorus         1.2                           1.2/100 = 0.012

Other elements   1.2                           1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

                           Percentage ( %)      Decimal number      Mass in grams (g)        

Oxygen               65.0                         65/100 = 0.65         0.65*62000=40300

Carbon               18.0                           18/100 = 0.18         0.18*62000=  11160

Hydrogen            10.0                         10/100 = 0.10         0.10*62000=   6200

Nitrogen              3.0                           3/100 = 0.03          0.03*62000= 1860

Calcium               1.6                           1.6/100 = 0.016        0.016*62000= 992

Phosphorus         1.2                           1.2/100 = 0.012        0.012*62000=744

Other elements   1.2                           1.2/100 = 0.012        0.012*62000=744  

5 0
3 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
Zn +<br> AgCI -<br> ZnCl2 +<br> - Ag<br><br> HELPPP?
garik1379 [7]

Answer:

Explanation:   Zn (s) + 2 AgCl (s) ⇒ ZnCl2 (s) + 2 Ag(s)

Less noble Zinc reduces more noble Silver

8 0
3 years ago
Can metals have covalent radius and van der Waal's radius instead of metallic radius?​
romanna [79]

Explanation:

Whether you choose to use van der Waals radii or metallic radii as a measure of the atomic radius, for metals the ionic radius is smaller than either, so the problem doesn't exist to the same extent. It is true that the ionic radius of a metal is less than its atomic radius (however vague you are about defining this).

4 0
3 years ago
Consider the equation for the formation of water.
Juli2301 [7.4K]

The theoretical yield is 160 g H₂O.

<em>Moles of H₂</em> = 18 g H₂ × (1 mol H₂/2.016 g H₂) = 8.93 mol H₂

<em>Moles of H₂O</em> = 8.93 mol H₂O × (2 mol H₂O/2 mol H₂) = 8.93 mol H₂O

<em>Theoretical yield</em> of H₂O = 8.93 mol H₂O × (18.02 g H₂O/1 mol H₂) = 160 g H₂O

6 0
3 years ago
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